Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 3}{x - 8} = \dfrac{7x + 11}{x - 8}$
Explanation: Multiply both sides by $x - 8$ $ \dfrac{x^2 + 3}{x - 8} (x - 8) = \dfrac{7x + 11}{x - 8} (x - 8)$ $ x^2 + 3 = 7x + 11$ Subtract $7x + 11$ from both sides: $ x^2 + 3 - (7x + 11) = 7x + 11 - (7x + 11)$ $ x^2 + 3 - 7x - 11 = 0$ $ x^2 - 8 - 7x = 0$ Factor the expression: $ (x - 8)(x + 1) = 0$ Therefore $x = 8$ or $x = -1$ At $x = 8$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 8$, it is an extraneous solution.